我正在嘗試使用執行程序服務來教自己使用多線程，并且想知道在下面實現我的代碼的最佳實踐是什么-我正在閱讀文本文件目錄并檢查字符/單詞-如果我所做的只是通過文件列表對每個線程進行處理，我也很困惑是否正在使用多個線程。是否同時處理多個文件？主班public class Application { private long totalCharacterCount; private long totalLineCount; private final File[] fileList; private final static String _DIRECTORY = "src//documents"; public Application(String directory){ fileList = new File(directory).listFiles(); } public synchronized File[] getFileList(){ return fileList; } public static void main(String[] args) throws InterruptedException, ExecutionException { ExecutorService executor = Executors.newFixedThreadPool(4); Application x = new Application(_DIRECTORY); for(File file : x.getFileList()){ Future<FileReadings> response = executor.submit(new Process(file)); x.totalCharacterCount += response.get().characterCount; x.totalLineCount += response.get().lineCount; } System.out.println("Total lines in all documents: " + x.totalLineCount); System.out.println("Total characters in all documents: " + x.totalCharacterCount); executor.shutdown(); }}進程類public class Process implements Callable<FileReadings> { private FileReadings object; private File file; public Process(File file){ FileReadings obj = new FileReadings(); this.object = obj; this.file = file; } public void CountCharacters(File file){ int count = 0; try { BufferedReader reader = Files.newBufferedReader(file.toPath()); while(reader.read() != -1){ count++; } object.characterCount = reader.read(); } catch (IOException ex) { ex.printStackTrace(); } object.characterCount = count; }
1 回答
慕田峪4524236
TA貢獻1875條經驗 獲得超5個贊
不。這不是正確的方法。您提交一個進程，然后通過調用get()未來，您阻塞并等待它完成，因此它實際上是一個同步處理。有兩種方法可以進行并行、異步處理：
1)invokeAll()
這是更簡單的方法，但它需要您提前創建所有流程實例，因此這取決于您要執行多少并行任務（如果您有數百萬個，您可能會達到內存限制）。創建流程后，您可以立即將它們提交給執行者。它將并行執行所有任務（根據線程池大?。┎⒃谒腥蝿胀瓿珊蠓祷亍?/p>
List<Callable<FileReadings>> tasks = new Arraylist<>();
for (File file : x.getFileList()) {
tasks.add(new Process(file));
}
// submit all processes at once. they will be processed in parallel
// this call blocks until all tasks are finished
List<Future<FileReadings>> responses = executor.invokeAll(tasks);
// at this point all processes finished. all get() will return immediately
for (Future<FileReadings> response : responses) {
x.totalCharacterCount += response.get().characterCount;
x.totalLineCount += response.get().lineCount;
}
2)submit()
當您創建一個進程并立即提交它時，此解決方案更具可擴展性，因此內存需求是恒定的（不包括執行程序）。但是，您需要自己管理響應：
List<Future<FileReadings>> responses = new ArrayList<>();
for (File file : x.getFileList()) {
responses.add(executor.submit(new Process(file)));
}
// at this point all processes submitted but not finished.
// need to check which is finished at intervarls
while (responses.isEmpty() == false) {
Thread.sleep(1000); // allow some processing time for tasks
// ListIterator allows removing items
ListIterator<Future<FileReadings>> itr = responses.listIterator();
while (itr.hasNext()) {
Future<FileReadings> response = itr.next();
// if task is complete, get it and remove from list
if (response.isDone()) {
x.totalCharacterCount += response.get().characterCount;
x.totalLineCount += response.get().lineCount;
itr.remove();
}
}
}
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